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標題:
中4 amaths 問題
發問:
1.Find the equations of the angle bisectors of the staight lines x-3y-6=0and 3x-y+4=02.Given sinx+2cosx=rsin(x+b),where r>0 and form0 0 and form0
最佳解答:
Q1. Suppose the slope of the equations of angle bisectors is m. Using tan x = l ( m1 - m2 ) / ( 1 + m1m2 ) l, we have l [ (1/3) - m ] / [ 1 + (m/3) ] l = l ( 1 - 3m ) / ( 1 + 3m ) l l 1 - 9m2 l = l 9 - m2 l 1 - 9m2 = 9 - m2 or m2 - 9 -8m2 = 8 ( rejected ) or 10m2 = 10 m = 1 or -1 Next, find out the intersection point of x - 3y - 6 = 0 and 3x - y + 4 = 0 On solving, it is ( -9/4, -11/4 ). Lastly, the required equations are 2x - 2y - 1 = 0 or x + y + 5 = 0 2a) sin x + 2 cos x = r sin ( x + b ) sin x + 2 cos x = r sin x cos b + r sin b cos x So, r cos b = 1 and r sin b = 2 Here we get r = sqr 5 and b = 63.4o b) 1/( sin x + 2 cos x + 3 sqr 5) = 1/[ sqr 5 sin ( x + 63.4o ) + 3 sqr 5 ] Max. value of y = 1 / ( -sqr 5 + 3 sqr 5 ) = 1 / 2 sqr 5 Min. value of y = 1 / ( sqr 5 + 3 sqr 5 ) = 1 / 4 sqr 5 c) ( sin x + 2 cos x ) / ( sin x + 2 cos x + 3 sqr 5 ) = sqr 5 sin ( x + 63.4o ) / [ sqr 5 sin ( x + 63.4o ) + 3 sqr 5 ] = 1 - 3 sqr 5 / [ sqr 5 sin ( x + 63.4o ) + 3 sqr 5 ] Max. value of z = 1 - ( 3 sqr 5 / 4 sqr 5 ) = 1 / 4 Min. value of z = 1 - ( 3 sqr 5 / 2 sqr 5 ) = - 1 / 2
其他解答:
中4 amaths 問題
發問:
1.Find the equations of the angle bisectors of the staight lines x-3y-6=0and 3x-y+4=02.Given sinx+2cosx=rsin(x+b),where r>0 and form0 0 and form0
最佳解答:
Q1. Suppose the slope of the equations of angle bisectors is m. Using tan x = l ( m1 - m2 ) / ( 1 + m1m2 ) l, we have l [ (1/3) - m ] / [ 1 + (m/3) ] l = l ( 1 - 3m ) / ( 1 + 3m ) l l 1 - 9m2 l = l 9 - m2 l 1 - 9m2 = 9 - m2 or m2 - 9 -8m2 = 8 ( rejected ) or 10m2 = 10 m = 1 or -1 Next, find out the intersection point of x - 3y - 6 = 0 and 3x - y + 4 = 0 On solving, it is ( -9/4, -11/4 ). Lastly, the required equations are 2x - 2y - 1 = 0 or x + y + 5 = 0 2a) sin x + 2 cos x = r sin ( x + b ) sin x + 2 cos x = r sin x cos b + r sin b cos x So, r cos b = 1 and r sin b = 2 Here we get r = sqr 5 and b = 63.4o b) 1/( sin x + 2 cos x + 3 sqr 5) = 1/[ sqr 5 sin ( x + 63.4o ) + 3 sqr 5 ] Max. value of y = 1 / ( -sqr 5 + 3 sqr 5 ) = 1 / 2 sqr 5 Min. value of y = 1 / ( sqr 5 + 3 sqr 5 ) = 1 / 4 sqr 5 c) ( sin x + 2 cos x ) / ( sin x + 2 cos x + 3 sqr 5 ) = sqr 5 sin ( x + 63.4o ) / [ sqr 5 sin ( x + 63.4o ) + 3 sqr 5 ] = 1 - 3 sqr 5 / [ sqr 5 sin ( x + 63.4o ) + 3 sqr 5 ] Max. value of z = 1 - ( 3 sqr 5 / 4 sqr 5 ) = 1 / 4 Min. value of z = 1 - ( 3 sqr 5 / 2 sqr 5 ) = - 1 / 2
其他解答:
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